The question is to determine the percentage rate of change of the function at t=1 and t=5 for:

$\displaystyle f(t)=e^{0.3t^2}$

I calculated the derivative to be:
$\displaystyle f'(t)=e^{0.3t^2}0.6t$

So is the answer to t=1, an increase of 60%?

And for t=5, 750%?

Nov 13th 2009, 04:31 AM

HallsofIvy

Quote:

Originally Posted by Kataangel

The question is to determine the percentage rate of change of the function at t=1 and t=5 for:

$\displaystyle f(t)=e^{0.3t^2}$

I calculated the derivative to be:
$\displaystyle f'(t)=e^{0.3t^2}0.6t$

So is the answer to t=1, an increase of 60%?

And for t=5, 750%?

For t= 1, yes, for t= 5, no.

The rate of increase is, as you say, the derivative, f'. The percentage rate of increase is that, divided by f: f'/f= [tex]\frac{e^{0.3t^2}0.6t}{e^{0.3t^2}}= 0.6t. For t= 1, that is .6(1)= .6 or 60%. For t= 5, that is .6(5)= 3.0 or 300%.